Skip to main content

Positions of Large Groups

Problem Description​

In a string s of lowercase letters, these letters form consecutive groups of the same character.

For example, a string like s = "abbxxxxzyy" has the groups "a", "bb", "xxxx", "z", and "yy".

A group is identified by an interval [start, end], where start and end denote the start and end indices (inclusive) of the group. In the above example, "xxxx" has the interval [3,6]. A group is considered **large** if it has3` or more characters.

Return the intervals of every large group sorted in increasing order by start index.

Examples​

Example 1:

Input: s = "abbxxxxzzy"
Output: [[3,6]]
**Explanation**: "xxxx" is the only large group with start index 3 and end index 6.

Example 2:

Input: s = "abc"
Output: []
**Explanation:** We have groups "a", "b", and "c", none of which are large groups.

Example 3:

Input: s = "abcdddeeeeaabbbcd"
Output: [[3,5],[6,9],[12,14]]
**Explanation:** The large groups are "ddd", "eeee", and "bbb".

Constraints​

  • 1 <= s.length <= 1000
  • s contains lowercase English letters only.

Solution for Assign Cookies​

Approach:​

1- For every groups, find its start index i and end index j - 1.

2- Group length is j - i, if it's no less than 3, add (i, j) to result.

Code in Different Languages​

C++​

class Solution {
    public:
        vector<vector<int>> largeGroupPositions(string S) {
            vector<vector<int>> res;
            for (int i = 0, j = 0; i < S.size(); i = j) {
                while (j < S.size() && S[j] == S[i]) ++j;
                if (j - i >= 3)
                res.push_back({i, j - 1});
            }
            return res;
            }
};

Java​

class Solution {
        public List<List<Integer>> largeGroupPositions(String S) {
        List<List<Integer>> res = new ArrayList<>();
        for (int i = 0, j = 0; i < S.length(); i = j) {
            while (j < S.length() && S.charAt(j) == S.charAt(i)) ++j;
            if (j - i >= 3)
                res.add(Arrays.asList(i, j - 1));
        }
        return res;
    }
}


Python​

class Solution(object):
        def largeGroupPositions(self, S):
        i, j, N = 0, 0, len(S)
        res = []
        while i < N:
            while j < N and S[j] == S[i]: j += 1
            if j - i >= 3: res.append([i, j - 1])
            i = j
        return res

Complexity Analysis​

Time Complexity: O(N)​

Space Complexity: O(N)​

References​